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Interactive Differential Equations: A Step-by-Step Approach to Methods & Modeling

Geoffrey W. Cox, Ph.D., ?

Section 2.6 Summary & Exercises

Summary of the Key Ideas.

  • What is a Solution?
    • A solution to a differential equation is a function that, when substituted into the equation, satisfies the equation.
  • Verifying Solutions
    • Verification of a solution involves substituting the function into the equation and simplifying to check if a true statement is obtained.
    • To verify a solution, all terms should ideally be moved to one side of the equation, simplifying the process.
  • Types of Solutions
    • Differential equations can have different types of solutions: general and particular.
    • The general solution includes arbitrary constants and represents a family of solutions.
    • A particular solution is obtained by assigning specific values to these constants.
  • Initial Conditions
    • Initial conditions are values given for the solution or its derivatives at a specific point, often used to find a particular solution.
    • They help specify which member of the general solution family is applicable in a specific context.

Exercises Exercises

1.

Discuss the difference between a general solution, family of solutions, and a particualr solution.

2.

Given the differential equation
\begin{equation*} y''-9y = 0 \end{equation*}
determine if the following functions are solutions of this differential equation.
a. \(\ds \,\, y = 3\)
b. \(\ds \,\, y = 0\)
c. \(\ds \,\, y = e^{3x}\)
d. \(\ds \,\, y = 3x\)
e. \(\ds \,\, y = 9e^x\)
f. \(\ds \,\, y = x^9\)
g. \(\ds \,\, y = 4e^{3x}\)
h. \(\ds \,\, y = e^{-3x}\)
i. \(\ds \,\, y = e^{3x} - 2e^{-3x}\)
Answer.
a. No
b. Yes
c. Yes
d. No
e. No
f. No
g. Yes
h. Yes
i. Yes

3.

Given the differential equation
\begin{equation*} y'' - 10y' + 25y = 0 \end{equation*}
determine if the following functions are solutions of this differential equation.
a. \(\ds \,\, y = 0\)
b. \(\ds \,\, y = x^2e^{5x}\)
c. \(\ds \,\, y = e^{5x}\)
d. \(\ds \,\, y = 5x\)
e. \(\ds \,\, y = xe^{5x}\)
f. \(\ds \,\, y = 5e^{5x}\)
g. \(\ds \,\, y = 1/25\)
h. \(\ds \,\, y = e^{-5x}\)
i. \(\ds \,\, y = (1+x)e^{5x}\)
Answer.
a. Yes
b. No
c. Yes
d. No
e. Yes
f. Yes
g. No
h. No
i. Yes

4.

Match each function on the left with a DE on the right if this function is a solution (i.e. is in the family of solutions) for this DE.
  1. \(\displaystyle y = 3x+x^2\)
  2. \(\displaystyle y = e^{-8x}\)
  3. \(\displaystyle y = \sin x\)
  4. \(\displaystyle y = x^{1/2}\)
  5. \(\displaystyle y = 6e^{8x}\)
  • \(\displaystyle \boxed{\phantom{D}} \quad 2x^2y'' + 3xy' = y\)
  • \(\displaystyle \boxed{\phantom{B}} \quad y'' + 10 y' + 16 y = 0\)
  • \(\displaystyle \boxed{\phantom{E}} \quad y' = 8 y\)
  • \(\displaystyle \boxed{\phantom{C}} \quad y'' + y = 0\)
  • \(\displaystyle \boxed{\phantom{A}} \quad 3y'' - 2y + \frac{4}{x}y = 12\)
Answer.
  • \(\displaystyle \boxed{D} \quad 2x^2y'' + 3xy' = y\)
  • \(\displaystyle \boxed{B} \quad y'' + 10 y' + 16 y = 0\)
  • \(\displaystyle \boxed{E} \quad y' = 8 y\)
  • \(\displaystyle \boxed{C} \quad y'' + y = 0\)
  • \(\displaystyle \boxed{A} \quad 3y'' - 2y + \frac{4}{x}y = 12\)

5.

Find all values of \(r\) such that \(y=e^{rx}\) is a solution to
\begin{equation*} y''' - 10y'' + 24y' = 0\text{.} \end{equation*}
Answer.
\(r = 0, 6, 4\)

6.

Find all values of \(r\) such that \(y = k e^{rx}\) is a solution to
\begin{equation*} y'' - 25y = 0\text{.} \end{equation*}
Answer.
\(r = -5, 5\)

7.

Find all values of \(r\) such that \(y(x) = c_1 e^{r x} + c_2\) is a solution to
\begin{equation*} 7y'' = 8y\text{.} \end{equation*}
Answer.
\(r = 8/7\)

8.

Find all values of \(r\) such that \(y = t^{r}\) is a solution to
\begin{equation*} t^{2}y''- 11ty' + 32y = 0 \quad (t>0)\text{.} \end{equation*}
Answer.
\(r = 4, 8\)

9.

Find all values of \(r\) such that \(y = rx^{2}\) is a solution to
\begin{equation*} y' = 5x\text{.} \end{equation*}
Answer.
\(r = 2.5\)

10.

Find all values of \(r\) such that \(y = \sin(rt)\) is a solution to
\begin{equation*} y'' + 9y = 0\text{.} \end{equation*}
Answer.
\(r = 2.5\)

11.

Find the value of the constant \(k\) in order for the differential equation
\begin{equation*} \frac{d^2y}{dt^2} - 8 \frac{dy}{dt} + ky = 0 \end{equation*}
to have the solution \(y = e^{3t}\text{.}\)
Answer.
\(k = 15\)

12.

Is \(\ds y = c_{1}e^{2x} + c_{2}xe^{2x}\) a solution of the differential equation
\begin{equation*} \ds \frac{d^{2}y}{dx^{2}} - 4\frac{dy}{dx} + 4y = 0? \end{equation*}
Answer.
Yes

Find the value of \(r\).

For each of the differential equations and function pairs, find all the possible values of \(r\) such that the function is a solution to the DE.
13.
\(y''' - 10y'' + 24y' = 0; \quad y=e^{rx}\)
Answer.
\(r = 0, 6, 4\)
14.
\(y'' - 25y = 0; \quad y = k e^{rx}\)
Answer.
\(r = -5, 5\)
15.
\(7y'' = 8y; \quad y(x) = c_1 e^{r x} + c_2\)
Answer.
\(r = 8/7\)
16.
\(t^{2}y''- 11ty' + 32y = 0; \quad y = t^{r} \quad (t>0)\)
Answer.
\(r = 4, 8\)
17.
\(y' = 5x; \quad y = rx^{2}\)
Answer.
\(r = 2.5\)
18.
\(y'' + 9y = 0; \quad y = \sin(rt)\)
Answer.
\(r = 2.5\)

19.

Explain the significance of initial condition(s) as they relate to the general solution of a differential equation.
Answer.
The initial condition(s) specify one or more points that the graph of the solution must pass through. This often allows one to solve for the constants in the general solution.
Therefore, the initial condition(s) act to reduce the family of solutions down to a smaller set of solutions or, ideally, a single particular solution.

20.

Does an initial condition always give you a particular solution?
Answer.
When there is one constant in the general solution, yes.
When there is more than one constant in the general solution, no.\(^{1}\)

21.

Consider the differential equation
\begin{equation*} x^2y' + xy = 1 \end{equation*}
  1. Verify that the family of functions
    \begin{equation*} y = \frac{\ln x + C}{x} \end{equation*}
    a solution to this differential equation.
  2. Find the particular solution that corresponds to the initial condition \(y(9)=8\text{.}\)
Answer.
Yes
\(C = 72-\ln(9) \approx 69.803\text{,}\) so \(\ds y = \frac{\ln x + 69.803}{x} \) is the particular solution

22.

Consider the differential equation
\begin{equation*} y^\prime = y-y^2 \end{equation*}
Which of the following is a family of solutions for this DE?
  1. \(\displaystyle \ds y=c\sin(-x)\)
  2. \(\displaystyle \ds y=\frac{1}{1+ce^{-x}}\)
  3. \(\displaystyle \ds y=\frac{1}{c+x^2}\)
Find the particular solution such that \(y(0) = 7\text{.}\)
Answer.
b. \(\ds y=\frac{1}{1+ce^{-x}}\)
\(c = \frac{1}{7}-1 \approx -0.857\text{,}\) so \(\ds y = \frac{1}{1+(-0.857)e^{-x}} \) is the particular solution
Solution.
Try \(y=\frac{1}{c+x^2}\) first.
Compute
\(y^\prime = \frac{-4x}{(c+x^2)^2}\)
\(y-y^2 = \frac{1}{c+x^2} - \frac{1}{(c+x^2)^2} = \frac{c+x^2}{(c+x^2)^2} - \frac{1}{(c+x^2)^2} = \frac{c+x^2-1}{(c+x^2)^2}\)
So \(y^\prime\neq y-y^2\) for this function, it is not the solution.
Now try \(y=\frac{1}{1+ce^{-x}}\text{.}\) Then
\(y^\prime = \frac{ce^{-x}}{1+ce^{-x}}\)
while
\(y-y^2=\frac{1}{1+ce^{-x}}-\frac{1}{(1+ce^{-x})^2} =\frac{1+ce^{-x}}{(1+ce^{-x})^2}-\frac{1}{(1+ce^{-x})^2} =\frac{ce^{-x}}{(1+ce^{-x})^2}\)
So this function is the solution
Setting \(y(0)=$y0\) we get
\($y0 = \frac{1}{1+c}\)
Solve for c, \(c = \frac{1}{$y0}-1=$c\)
So the solution for the IVP is \(y = \frac{1}{1+(-0.857)e^{-x}}\)

Understanding what it means to Solve a DE.

We spend a lot of time solving equations, but do you even remember what it means to be a solution? Let’s review what it means to be a solution to an algebraic equation (as in question (8)), and then what it means to be a solution to a differential equation (as in question (9)), and then we’ll extend those ideas in questions (10) and (11).
23.
Verify that \(x = 3\) is a solution to the equation
\begin{equation} 2x^2 + 3 = 7x.\tag{8} \end{equation}
What confirms that \(x = 3\) is a solution?
Answer.
the LHS = RHS after substituting \(x = 3\) into both sides of the equation.
(Note: Do not rearrange the equation to solve for \(x\text{.}\) That would be the process of finding a solution; you are asked to verify a solution.)
Answer.
Since we get the same result (21) when we substitute the solution into the RHS and into the LHS, we confirm that \(x = 3\) is a solution to the equation.
Solution.
We substitute \(x = 3\) into the left hand side (LHS) of the equation, and then into the right hand side (RHS) of the equation:
\begin{align*} \text{LHS} =\amp\ 2x^2 + 3 \\ =\amp\ 2(3)^2 + 3 \\ =\amp\ 18 + 3 \\ =\amp\ 21 \\ \amp \\ \text{RHS} =\amp\ 7x \\ =\amp\ 7(3) \\ =\amp\ 21 \end{align*}
Since we get the same result when we substitute the solution into the RHS and into the LHS, we have confirmed that \(x = 3\) is a solution to the equation.
24.
Verify that \(y = c_1 \sin x + c_2 \cos x\) is a solution to the DE
\begin{equation} \ds \frac{d^2y}{dx^2} + y = 0.\tag{9} \end{equation}
Hint.
Look back at the previous question and think about what it means to be a solution to a DE.
Answer.
Since we get the same result (0) when we substitute the solution into the RHS and into the LHS, we confirm that \(y = c_1 \sin x + c_2 \cos x\) is a solution to the DE.
Solution.
We need to substitute into the DE. In order to substitute into the left hand side, we need to know the second derivative. So we begin by finding that.
\begin{align*} y =\amp\ c_1 \sin x + c_2 \cos x \\ \Rightarrow \quad y' =\amp\ c_1 \cos x - c_2 \sin x \\ \Rightarrow \quad y'' =\amp\ -c_1 \sin x - c_2 \cos x \end{align*}
Now we can substitute into the DE:
\begin{align*} \text{LHS}=\amp\ \frac{d^2y}{dx^2} + y \\ =\amp\ \Big[ -c_1 \sin x - c_2 \cos x \Big] + \Big[ c_1 \sin x + c_2 \cos x \Big] \\ =\amp\ 0 \\ =\amp\ \text{RHS} \end{align*}
Since when we substitute in to the DE, we find that the LHS is equal to the RHS, we have verified that \(y = c_1 \sin x + c_2 \cos x\) is a solution to the DE \(\ds \frac{d^2y}{dx^2} + y = 0.\)
25.
Consider the DE
\begin{equation} \ds \frac{d^2 y}{dx^2} + 6 \frac{dy}{dx} + 5y = 0.\tag{10} \end{equation}
Find all values of \(m\) such that \(y = e^{mx}\) is a solution to the DE by taking derivatives and substituting into the DE, and then solving for \(m\text{.}\)
Hint.
\(e^{mx} \ne 0,\) no matter the value of \(x\) or \(m.\) Therefore we might want to factor it out as it won’t contribute to our answer.
Answer.
\begin{equation*} m = -5, -1 \end{equation*}
Solution.
First we take derivatives so that we can substitute into the DE.
\begin{align*} y =\amp\ e^{mx} \\ \Rightarrow \quad y' =\amp\ me^{mx} \\ \Rightarrow \quad y'' =\amp\ m^2 e^{mx} \end{align*}
Now we can substitute into the DE and solve for \(m\text{:}\)
\begin{align*} \frac{d^2 y}{dx^2} + 6 \frac{dy}{dx} + 5y =\amp\ 0 \\ \Big[ m^2 e^{mx} \Big] + 6\Big[ m e^{mx} \Big] + 5\Big[ e^{mx} \Big] =\amp\ 0 \\ e^{mx} \Big[ m^2 + 6m + 5 Big] =\amp\ 0 \\ e^{mx} (m+5)(m+1) =\amp\ 0 \end{align*}
Recall that \(e^{mx} \ne 0,\) no matter the value of \(x\) or \(m.\) Thus we have:
\begin{align*} m+5 = 0 \amp \text{ or } m+1 = 0 \\ m = -5, \amp \text{ or } m = -1 \end{align*}
So the two solutions are \(m = -5, -1.\)
26.
Now that you have (hopefully!) two values for \(m,\) choose one of the two values and then verify that \(y = e^{mx}\) is a solution for that value of \(m.\)
Answer.
Since we get the same result (0) when we substitute the solution into the RHS and into the LHS, we confirm that \(y = e^{-5x}\) (or \(y = e^{-x}\)) is a solution to the DE.
Solution.
The work here is for \(m = -5,\) and the work for \(m = -1\) is similar.
First we take derivatives so that we can substitute into the DE.
\begin{align*} y =\amp\ e^{-5x} \\ \Rightarrow \quad y' =\amp\ -5e^{-5x} \\ \Rightarrow \quad y'' =\amp\ 25 e^{-5x} \end{align*}
Now we can substitute into the DE:
\begin{align*} \text{LHS}=\amp\ \frac{d^2 y}{dx^2} + 6 \frac{dy}{dx} + 5y \\ =\amp\ \Big[ 25e^{-5x} \Big] + 6\Big[ -5e^{-5x} \Big] + 5\Big[ e^{-5x} \Big] \\ =\amp\ 25e^{-5x} - 30 e^{-5x} + 5e^{-5x} \\ =\amp\ 0 \\ =\amp\ \text{RHS} \end{align*}
Hence, we have shown that \(y = e^{-5x}\) is a solution to the DE \(\ds \frac{d^2 y}{dx^2} + 6 \frac{dy}{dx} + 5y = 0.\)
27.
Assuming the form of the solution of DE.
Find all values of \(m\) such that \(y = x^m\) is a solution of the DE
\begin{equation} \ds 3x^2 \frac{d^2y}{dx^2} = -11x \frac{dy}{dx} + 3y.\tag{11} \end{equation}
Answer.
\begin{equation*} m = -3, \frac{1}{3} \end{equation*}
Solution.
We proceed as in the previous question. First we take derivatives so that we can substitute into the DE.
\begin{align*} y =\amp\ x^m \\ \Rightarrow \quad y' =\amp\ mx^{m-1} \\ \Rightarrow \quad y'' =\amp\ m(m-1)x^{m-2} \end{align*}
Now we can substitute into the DE:
\begin{align*} 3x^2 \frac{d^2y}{dx^2} =\amp\ -11x \frac{dy}{dx} + 3y \\ 3x^2\Big[ m(m-1)x^{m-2} \Big] =\amp\ -11x\Big[ mx^{m-1} \Big] + 3\Big[ x^m \Big] \\ 3m(m-1)x^2 x^{m-2} =\amp\ -11mx^1\cdot x^{m-1} + 3x^m \\ (3m^2 - 3m)x^{2+(m-2)} =\amp\ -11m x^{1 + (m-1)} + 3x^m \\ (3m^2 - 3m)x^{m} =\amp\ -11m x^{m} + 3x^m \\ (3m^2 - 3m)x^{m} + 11m x^{m} - 3x^m =\amp\ 0 \\ (3m^2 - 3m + 11m - 3)x^m =\amp\ 0 \\ (3m^2 +8m - 3)x^m =\amp\ 0 \\ (3m -1)(m+3)x^m =\amp\ 0 \end{align*}
Note that there is no value of \(m\) that can guarantee that \(x^m\) is 0. Hence we have
\begin{align*} 3m-1 = 0 \amp \quad\text{or}\quad m+3 = 0 \\ m = \frac{1}{3}, \amp \quad\text{or}\quad m = -3 \end{align*}
So the two solutions are \(m = -3, \frac{1}{3}.\)
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